Question

[JEE Main-2021] 2. If the distance of the point \( (1,-2,3) \) from the plane \( x+2 y-3 z+10=0 \) measured parallel to the line, \( \frac{x-1}{3}=\frac{2-y}{m}=\frac{z+3}{1} \) is \( \sqrt{\frac{7}{2}} \), then the value of \( | m | \) is equal to [JEE Main-2021]

Answer 1

Equation of the line passing through P( 1, −2, 3 ) parallel to line

 (x - 1)/3 =  (2 − y)/m =  (z + 3)/1   

is 

 (x - 1)/3 =  (y + 2)/-m =  (z - 3)/1

let general point on above line Q( 3t+1, -mt-2 , t+3 )

Point Q lines on a plane x + 2y - 3z + 10 = 0
then  3t+1 + 2(-mt-2 ) - 3( t + 3 ) + 10 = 0
-2mt-2 = 0
then  t = -1/m

Q (  (-3/m)+1, -1, (-1/m) + 3 )

given distance between P and Q = sqrt[7/2]

distance between P and Q = sqrt[ (3/m)² + (-1)² + (1/m)² ]
sqrt [10/(m)² + 1]

then sqrt [10/(m)² + 1] = sqrt[7/2]
10/(m)²  = 5/2

|m| = 2